An offbeat question involving Milnor's $K_2$ has come up recently. Start with an algebraically closed field $F$ (perhaps required to be of characteristic 0). Let $G$ be a connected, simply connected simple algebraic group over $F$, for instance $\mathrm{SL}_n(F)$, maybe of rank $\neq 1,2$ to be on the safe side. Then consider a linear group $H \subset \mathrm{GL}(V)$ with $V$ finite dimensional over $F$, together with an epimorphism $\pi:H \rightarrow G$ of abstract groups whose kernel lies in the center of $H$. (EDIT: Further assume that $H$ is equal to its derived group. I neglected to include this crucial condition in the question originally posed to me.)

Are these conditions enough to imply that $\pi$ has trivial kernel?

Note that when $G$ is a special linear group of rank at least 2, its abstract universal central extension is the Steinberg group (with generators and relations specified by Matsumoto) and the kernel of the resulting map is $K_2(F)$. Typically this is an infinite group, uncountable if $F$ is uncountable (Milnor). However, if we add the assumption that $H$ acts irreducibly on $V$, then by Schur's Lemma $\pi$ induces a sort of reverse projective representation of $G$ with image equal to the image of $H$ in $\mathrm{PGL}(V)$. Older work of Steinberg would allow us to lift this projective representation to an ordinary one, thus mapping $G$ onto $H$. (See sections 6,7 of Steinberg's Yale lectures here.)

However, it's apparently undesirable in the original question to make any assumption about complete reducibility of the action of $H$. So I'm unsure what can be said, given only that $H$ is a linear group.